(c) `(7^2)^3 /7^3 =`

1.Solve the following : (a) (3/7)^8/(3/7)^2

Evaluate int ((8x+13)/sqrt(4x+7)) dx (A) 1/3 (4x+7)^(3/2) - 1/2 (4x+7)^(1/2) + c (B) 1/6 (4x+7)^(5/2) - 2/3 (4x+7)^(3/2) + c (C) 1/3 (4x+7)^(5/2) - 1/2 (4x+7)^(3/2) + c (D) (4x+7)^(3/2) - 1/2 (4x+7)^(1/2) + c

If projections of as line on x,y and z axes are 6,2 and 3 respectively, then directions cosines of the lines are (A) (6/2,2/7,3/7) (B) (3/5,5/7,6/7) (C) (1/7,2/7,3/7) (D) none of these

(2/3)^(-5)x\ (5/7)^(-5) is equal to (a) (2/3x5/7)^(-10) (b) (2/3x5/7)^(-5) (c) (2/3x5/7)^(25) (d) (2/3x5/7)^(-25)

Express the following as improper fractions : (a) (7) 3/4 (b) (5) 6/7 (c) (2) 5/6 (d) (10) 3/5 (e) (9) 3/7 (f) (8) 9/4

If in A B C ,A=pi/7,B=(2pi)/7,C=(4pi)/7 then a^2+b^2+c^2 must be R^2 (b) 3R^2 (c) 4R^2 (d) 7R^2

If in A B C ,A=pi/7,B=(2pi)/7,C=(4pi)/7 then a^2+b^2+c^2 must be R^2 (b) 3R^2 (c) 4R^2 (d) 7R^2

The direction cosines of a normal to the plane 2x-3y-6z+14=0 are (A) (2/7,(-3)/7,(-6)/7) (B) ((-2)/7,3/7,6/7) (C) ((-2)/7,(-3)/3,(-6)/7) (D) none of these

In a triangle A B C ,a=4,b=3,/_A=60^0 then c is root of the equation c^2-3c-7=0 (b) c^2+3c+7=0 (c) c^2-3c+7=0 (d) c^2+3c-7=0

If a=(1)/(7)(2hat(i)+3hat(j)+6hat(k)): b=(1)/(7)(6hat(i)+2hat(j)-3hat(k)): c=c_1hat(i)+c_2hat(j)+c_2hat(k) and matrix A=[[(2)/(7), (3)/(7), (6)/(7)], [(6)/(7), (2)/(7), -(3)/(7)], [c_1, c_2, c_3]] and AT^(T)=I , then c